QUESTIONS, HINTS & SOLUTIONS MATHEMATICS PAPER-1 · 2020. 11. 28. · PAPER-1 SECTION Œ 1 :...

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QUESTIONS, HINTS & SOLUTIONS

MATHEMATICS PAPER-1

SECTION � 1 : (Maximum Marks : 27)

This section contains TWELVE (12) questions. Each question has FOUR options (A), (B), (C) and (D) ONLY ONE of these four option is correct There are 12 Questions & you have attempt any 9 Questions. If a student attempts more than 9

questions, then only first 9 questions which he has attempted will be checked. Marking scheme :

Full Marks : +3 If ONLY the correct option is chosen. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : �1 In all other cases

[kaM 1 : (vf/kdre vad : 27) bl [kaM esa ckjg (12) iz'u gSaA

izR;sd iz'u esa pkj fodYi (A), (B), (C) rFkk (D) gSaA bu pkj fodYiksa esa ls dsoy ,d fodYi lgh gSaA

bl [kaM esa 12 iz'u gaS ftuesa ls vkidks dsoy fdUgh 9 iz'uksa dk mÙkj nsuk gS ;fn vki 9 ls vf/kd iz'uksa dk mÙkj nsrs gS] rks mÙkj

fn;s x;s izFke 9 iz'uksa dh gh tk¡p dh tk;sxhA vadu ;kstuk :

iw.kZ vad % +3 ;fn flQZ lgh fodYi gh pquk x;k gSA

'kwU; vad % 0 ;fn dksbZ Hkh fodYi ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A _.k vad % �1 vU; lHkh ifjfLFkfr;ksa esaA

1. If y = sinx

| x |log

x

, then the possible set of values of x and y are

;fn y = sinx

| x |log

x

gks] rks x rFkk y ds ekuksa dk lEHkkfor leqPp; gS

(A) x n W

(2n , 2n + )

and vkSj y {0}

(B) x (0, ) and vkSj y {1}

(C*) x n W

2n , 2n 2n , (2n 1)2 2

and vkSj y {0}

(D) x n W

(2n , (2n +1) )

and vkSj y {0, 1}

(Where W stands for the set of all the whole numbers) (tgk¡ W iw.kZ la[;kvksa ds leqPp;ksa dks crkrk gS½

Sol. sin x

| x |log

x sin x (0, 1) and x (0, )

x n W

2n , 2n 2n , (2n 1)2 2

and y {0}

ADVANCED PATTERN PART TEST-2(APT-2)

TARGET : JEE (MAIN+ADVANCED) 2021

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mailto:[email protected]

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2. Consider the following statements :

S1 : If x > 0, then x + 1x

2.

S2 : Minimum value of 9 tan2 + 4cot2 is 12.

S3 : If [x] and {x} represent integral and fractional parts of x, then the expression [x] + 2000

r 1

{x r}2000

= x.

State, in order, whether S1, S2, S3 are true or false fuEufyf[kr dFkuksa ij fopkj dhft, %

S1 : ;fn x > 0, rc x + 1x

2

S2 : 9 tan2 + 4cot2 dk U;wure eku 12 gSA

S3 : ;fn [x] vkSj {x}, x ds iw.kk±d o fHkéka'k Hkkx dks O;ä djrk gS] rks O;atd [x] + 2000

r 1

{x r}2000

= x gSA

S1, S2, S3 ds lR; (T) ;k vlR; (F) gksus dk lgh Øe gS & (A*) TTT (B) TTF (C) TFT (D) TFF Sol. S1 : Using A.M. G.M.

x > 0 x + 1x

2. True.

S2 : Using A.M. G.M. 2 29 tan 4cot

2

6.

Minimum value of 9tan2 + 4cot2 = 12

S3 : L.H.S = [x] + 2000

r 1

{x r}2000

= [x] + 2000

r 1

{x}2000

= [x] + {x} = x Hindi S1: A.M. G.M. ls x > 0 x + 2 lR; gS

S2 : A.M. G.M. ls 2 29 tan 4cot

2

6

9tan2 + 4cot2 = 12 dk U;wure eku

S3 : L.H.S = [x] + 2000

r 1

{x r}2000

= [x] + 2000

r 1

{x}2000

= [x] + {x} = x

3. Let f : R R satisfy relation f(x) f(y) � f(xy) = x + y x, y R and f(1) > 0. If h(x) = f(x) f �1(x), then

length of longest interval in which h(sinx+cosx) is strictly decreasing is. ekuk f : R R lEcU/k f(x) f(y) � f(xy) = x + y x, y R rFkk f(1) > 0 dks larq"V djrk gSA ;fn h(x) = f(x) f

�1(x), rc vf/kdre vUrjky dh yEckbZ ftlesa h(sinx+cosx) fujUrj áleku gS&

(A) (B*) 2

(C) 4

(D) 6

Sol. f(x).f(y)�f(xy) = x + y ...(1) put x = y = 1 j[kus ij f ²(1)�f(1) = 2 f ²(1) �f(1) � 2 = 0



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f(1) = 2 or �1 f(1) = 2 ( f(1) > 0)

Now put y = 1 in eqn. (1) lehjd.k (1) esa y = 1 j[kus ij f(1).f(x) � f(x) = x + 1 f(x) = x + 1 y = x+ 1 x = y � 1

f�1(x) = x � 1

f(x). f�1(x) = x² � 1 = h(x) h(sinx+ cosx) = (sinx + cosx)² � 1 = sin2x

Required interval length is vf/kdre vUrjky dh yEckbZ 2

.

4. If f() =

2

2

cos cos sin sin

cos sin sin cos

sin cos 0

, then f3

+ f23

+ f() + f43

+ ......... + f n3

=

(A*) n (B) n(n 1)

2

(C) n2 + 2n (D) 2n2 � n

;fn f() =

2

2

cos cos sin sin

cos sin sin cos

sin cos 0

gS] rks f3

+ f23

+ f() + f43

+ ......... + f n3

=

(A*) n (B) n(n 1)

2

(C) n2 + 2n (D) 2n2 � n

Sol. C1 C1 � sin C3 ; C2 C2 � cos C3

f() =

1 0 sin

0 1 cos

sin cos 0

= 1.(0 + cos2 ) � 0 � sin (�sin ) = 1

f 3

= 1, f 23

= 1,..........., f n3

= 1

Given expression = n ¼fn;k O;atd = n½

5. Let A = {1, 2, 3, 4} and B = {a, b, c, d} The number of into functions from A to B is ekukfd A = {1, 2, 3, 4} rFkk B = {a, b, c, d} gS] rks A ls B esa cuus okys vUr{ksZih Qyuksa dh la[;k gS & (A*) 232 (B) 164 (C) 128 (D) 64 Sol. Total number of function = 44 = 256 Number of onto functions = 4! = 24 Number of into function = 232 Hindi. Qyuksa dh dqy la[;k = 44 = 256 vkPNknd Qyuksa dh la[;k = 4! = 24 vUr{ksZih Qyuksa dh la[;k = 232

6. The solution set of the equation sin1 tan4

� sin1 3x

� 6

= 0 is

lehdj.k sin1 tan4

� sin1 3x

� 6

= 0 dk gy leqPp; gS&

(A) { 2 } (B) {� 4 } (C*) { 4 } (D) {�2 }



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Sol. sin�1 �1 3tan � sin �

4 x 6

= 0

sin�1 1 � sin�1 3x

= 6

3

2 =

3x

x = 2 x = 4

7. The interval on which cos�1 x > sin�1 x > tan�1 x, is cos�1 x > sin�1 x > tan�1 x ds fy, vUrjky gS&

(A*) 1

0,2

(B) [�1, 1] (C) (0,1] (D) 1

�1,2

Sol. cos�1 x > sin�1 x > tan�1 x is defined for x [�1, 1] ; Now cos�1 x > sin�1 x cos�1 x > sin�1 x > tan�1 x , x [�1, 1] ds fy, ifjHkkf"kr gS vc cos�1 x > sin�1 x

cos�1 x > 2� cos�1 x

2cos�1 x > 2

cos�1 x > 4

cos�1 x ,4

x [cos , cos /4) as cos�1 x is a decreasing and continuous function x [cos , cos /4) as cos�1 x áleku Qyu rFkk lrr~ Qyu gSA

x [�1, 1/ 2 ) Also from the graph given below, it is clear that sin�1 x > tan�1 x for x (0, 1] in the domain [�1,1] fn, x;s vkjs[k ls uhps ;g Li"V gS fd sin�1 x > tan�1 x, x (0, 1] ds fy, izkUr [�1,1]

Thus cos�1 x > sin�1 x for x 1

1,2

and sin�1 x > tan�1 x for x (0,1]

vr% x 1

1,2

ds fy, cos�1 x > sin�1 x vkSj x (0,1] ds fy, sin�1 x > tan�1 x

cos�1 x > sin�1 x > tan�1 x for x 1 1

1, (0,1] 0,2 2

x 1 1

1, (0,1] 0,2 2

ds fy, cos�1 x > sin�1 x > tan�1 x

8. If y = f(x3), z = g(x5), f '(x) = tanx and g '(x) = sec x. Then value of x 0

(dy / dz)lim

x is

;fn y = f(x3), z = g(x5), f '(x) = tanx vkSj g '(x) = sec x rc x 0

(dy / dz)lim

x dk eku gS &

(A) 25

(B) 1 (C*) 35

(D) 45



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sol. 3 2 3

5 4 5 2

dy f '(x )3x tan x 3.

dz g'(x ).5x sec x 5x ; Hence vr%

3

2 5x 0

3 tanx 1 3lim .

x 55x .sec x

9. cot�1 92

+ cot�1 334

+ cot�1 129

8

............ dk eku gS&

(A) 4

(B) 2

(C) 4

(D*) buesa ls dksbZ ugha

Sol. 1 1 1 1 19 2 4 2cot tan tan tan 4 tan 2

2 9 1 4.2

1 1 1 1 133 4 8 4cot tan tan tan 8 tan 4

4 33 1 4.8

1 1 1 1 1129 8 16 8cot tan tan tan 16 tan 8

8 129 1 16.8

blh izdkj Tn = 1 n 1 1 ntan 2 tan 2

dks tksMUks ij

;ksx = 1 n 1 1tan 2 tan 2

tc n , rc ;ksx = 2

� tan�1 2 = cot�1 2.

10. If a curve is represented parametrically by the equations x = 4t3 + 3 and y = 4 + 3t4 and

2

2

n

d x

dy

dxdy

is

constant then value of n is

;fn ,d oØ dh izkpfyd lehdj.ksa x = 4t3 + 3 vkSj y = 4 + 3t4 gS rFkk

2

2

n

d x

dy

dxdy

vpj gS rc n dk eku gS&

(A) 3 (B) 4 (C*) 5 (D) 6

Sol. 2

3

dxdx 12t 1dt

dydy t12tdt

2

2 2 3 5

d x d dx d 1 dt 1 1 1. .

dy dy dt t dydy t 12t 12t

So 5

n

1

12t

1t

is constant n = 5

blfy, 5

n

1

12t

1t

vpj gSA n = 5

11. Let f(x) be a polynomial in x. Then the second derivative of f(ex) w.r.t. x is

;fn f(x), x esa ,d cgqin gS rc f(ex) dk x ds lkis{k f}rh; vodyt gS &

(A) f (ex). ex + f (ex) (B*) f (ex). e2x + f (ex). ex

(C) f (ex) e2x (D) f (ex). ex + f (ex). e2x



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Sol. xd(f(e ))

dx = f(ex).ex

2

x2

d(f(e ))

dx = ex f(ex) + e2x f(ex)

12. 2

2

d x

dy =

(A*) � 2

2

d y

dx

3dy

dx

(B)

12

2

d y

dx

(C) �

12

2

d y

dx

3dy

dx

(D) 2

2

d y

dx

2dy

dx

Sol. dy 1

dxdxdy

2

2

d y d 1

dx dx / dydx

=

d 1 dy.

dy dx / dy dx

= �

2

2

2

d x

1 dy.

dxdxdydy

=

2

2

3

d x�

dy

dxdy

SECTION � 2 : (Maximum Marks : 18) This section contains SIX (06) questions. The answer to each question is NUMERICAL VALUE with two

digit integer and decimal upto two digit. If the numerical value has more than two decimal places truncate/round-off the value to TWO decimal

placed. Marking scheme :

Full Marks : +3 If ONLY the correct option is chosen.

Zero Marks : 0 In all other cases

[kaM 2 ¼vf/kdre vad% 18) bl [kaM esa N% (06) iz'u gSA izR;sd iz'u dk mÙkj la[;kRed eku (NUMERICAL VALUE) gSa]tks f}&vadh; iw.kkZad

rFkk n'keyo f}&vadu eas gSA ;fn la[;kRed eku esa nks ls vf/kd n’'keyo LFkku gS ] rks la[;kRed eku dks n'keyo ds nks LFkkuksa rd VªadsV@jkmaM

vkWQ (truncate/round-off) djsaA vadu ;kstuk %&

iw.kZ vad % +3 ;fn flQZ lgh la[;kRed eku (Numerical value) gh mÙkj Lo:i ntZ fd;k x;k gSA 'kwU; vad % 0 vU; lHkh ifjfLFkfr;ksa esaA

13. If f(x) = tan�1 3sin x4 5cosx

, then find the value of 7f

3

.

;fn f(x) = tan�1 3sin x4 5cosx

gS] rks 7f

3

dk eku Kkr dhft,A

Ans. 03.00

Sol. f (x) = 2 2

1 3cosx(4 5cos x) � 3sinx(�5sinx)

(4 5cosx)3sinx1

4 5cos x

f (x) =

2 2 2

2 22

(4 5cosx) 12cosx 15cos x 15sin x(4 5cosx)(4 5cos x) 3sinx



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f (x) = 2 2

3(4cosx 5)16 40cos x 25cos x 9(1� cos x)

f (x) = 2

3(4cosx 5)(4cosx 5)

=

34cosx 5

f 3

= 37

14. If f(x) = max x

,| sin x |n

, n N has maximum points of non-differentiability for x (0, 4) then what is

the least value of n.

;fn f(x) = max x

,| sin x |n

, n N, x (0, 4) ds fy, vodyuh;rk ugha gksus ds vf/kdre fcUnqvksa dks j[krk

gS] rc n dk U;wure eku gS&

Ans. (4)

Sol.

1 2 3 4 3.5

f(x) = max x

,| sin x |n

for max points vf/kdre fcUnq ds fy,

n > 3.5

nmin = 4

15. If 1 2 1 2 1 2 1

limx 0

sin x 2 sin 2x 3 sin 3x n sin nx...... 2025

x x x x

then find the value of 'n'

(where [.] denotes the greatest integer function)

;fn

1 2 1 2 1 2 1limx 0

sin x 2 sin 2x 3 sin 3x n sin nx...... 2025

x x x x

rc 'n' dk eku Kkr dhft,A

(tgk¡ [.] egÙke iw.kkZad Qyu dks O;Dr djrk gSA)

Ans. 09.00

Sol. In vicinity of 1

1 sin xx 0, sin x | x | 1,

x

in vicinity of x 0

x = 0 ds fy,

11 sin x

sin x | x | 1x

vkSj x 0 ds fy,

3 3 3 3l 1 2 3 ...................... n 2025

1 3 1 3 1 3 1

limx 0

sin x 2 sin 2x 3 sin x n .sin nx............

x 2x 3x nx

2

2n(n 1)2025 (45)

2

n(n 1)

452

2n n 90 0 (n 10)(n 9) 0 n 9

16. If f(x) = x2 � 3x + 1 ; x 2 and g(x) is inverse function of f(x), then find the value of 1g (1)

.



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;fn f(x) = x2 � 3x + 1 ; x 2 vkSj g(x), f(x) dk izfrykse Qyu gS] rc 1

g (1) dk eku Kkr dhft,A

Ans. 03.00 Sol. f(g(x)) = x

f (g(x)). g (x) = 1

g (x) = 1

f (g(x))

g (1) = 1

f (g(1)) [where tgk¡ f(x) = x2 � 3x + 1 = 1, x = 0, x = 3 but ijUrq x 2]

= 1

f (3)

g (1) = 13

1

3g (1)

17. 2

x2

1 sinx sin3x sin5x.sin7xlim

x2

is k then k6

equal to

2

x2

1 sinx sin3x sin5x.sin7xlim

x2

dk eku k gS rks k6

cjkcj gS �

Ans. 07.00

Sol. x � 2

= y

2y 0

1 cosycos3ycos5ycos7ylim

y

y 0

sinycos3ycos5ycos7y 3cos ysin3ycos5ycos7y 5cosycos3ysin5ycos7y 7cos ycos3ycos5ysin7ylim

2y

1 9 25 4942

2 2 2 2

18. The domain of the function f(x) = x 1 x3 2 2

+ 1sin x

is [a, b], then find the value of a2 + b2 + ab.

Qyu f(x) = x 1 x3 2 2

+ 1sin x

dk izkUr [a, b] gS, rc a2 + b2 + ab dk eku Kkr dhft,A Ans. 01.00

Sol. Letekuk (x) = x 1 x3 2 2

andvkSj (x) = 1sin x

So vr% f(x) = (x) + (x) Letekuk dom = D1 dom = D2 . Then rc dom f = D1 D2 Sovr%, D1 = {x R | 3 � 2x � 21 � x 0] and vkSj D2 = {x R | sin�1 x 0} Nowvc, 3 � 2x � 21 � x 0

or ;k 3 � 2x � x

22

0

or ;k 3 . 2x � (2x)2 � 2 0 or ;k (2x)2 � 3.2x + 2 0



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(2x � 2) (2x � 1) 0 By sign-scheme, fpUg ifjorZu fof/k ls

2x [1, 2], i.e., 20 2x 2; 0 x 1 D1 = [0, 1]

Nowvc, sin�1 x 0 0 sin �1 x 2

{ the principal value of sin�1x is positive in the first quadrant} { sin�1x dk eq[; eku izFke prqFkk±'k esa /kukRed gS } 0 x 1 D2 = [0, 1] dom f = D1 D2 = [0, 1] [0, 1] = [0, 1] a = 0 andvkSj b = 1 a2 + b2 + ab = 1


This section contains TWO (02) paragraphs Based on each paragraph, there will be TWO questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s)

is(are) correct. For each question, marks will be awarded in one of the following categories :

Full Marks : +4 If only (all) the correct option(s) is (are) chosen. Partial Marks : +3 If all the four options are correct but ONLY three options are chosen. Partial Marks : +2 If three or more options are correct but ONLY two options are chosen

and both of which are correct. Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and

it is a correct option. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : �2 In all other cases.

For example,in a question, if (A),(B) and (D) are the ONLY three options corresponding to correct answers, then : Choosing ONLY (A),(B) and (D) will get +4 marks Choosing ONLY (A) and (B) will get +2 marks Choosing ONLY (A) and (D) will get +2 marks Choosing ONLY (B) and (D) will get +2 marks Choosing ONLY (A) will get +1 marks Choosing ONLY (B) will get +1 marks Choosing ONLY (D) will get +1 marks Choosing no option (i.e. the question is unanswered) will get 0 marks, and Choosing any other combination of options will get �2 mark

[kaM 3 : (vf/kdre vad : 16) bl [kaM esa nks (02) vuqPNsn gSaA izR;sd vuqPNsn ij nks iz'u gaSA

izR;sd iz'u esa pkj fodYi (A), (B), (C) rFkk (D) gSaA bu pkj fodYiksa esa ls ,d ;k ,d ls vf/kd fodYi lgh gSaA

izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk;saxs %

iw.kZ vad % +4 ;fn dsoy ¼lkjs½ lgh fodYi ¼fodYiksa½ dks pquk x;k gSA

vkaf'kd vad % +3 ;fn pkjksa fodYi lgh gSa ijUrq dsoy rhu fodYiksa dks pquk x;k gSA

vkaf'kd vad % +2 ;fn rhu ;k rhu ls vf/kd fodYi lgh gS ijUrq dsoy nks fodYiksa dks pquk x;k gS vkSj



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nksauks pqus gq, fodYi lgh fodYi gSaA

vkaf'kd vad % +1 ;fn nks ;k nks ls vf/kd fodYi lgh gSa ijUrq dsoy ,d fodYi dks pquk x;k gS vkSj

pquk gqvk fodYi lgh fodYi gSA

'kwU; vad % 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A

_.k vad % �2 vU; lHkh ifjfLFkfr;ksa esaA

mnkgj.k% ;fn fdlh iz'u ds fy, dsoy fodYi (A),(B) vkSj (D) lgh fodYi gS] rc %

dsoy fodYi (A),(B) vkSj (D) pquus ij +4 vad feysaxs ’ dsoy fodYi (A) vkSj (B) pquus ij +2 vad feysaxs ’ dsoy fodYi (A) vkSj (D) pquus ij +2 vad feysaxs ’ dsoy fodYi (B) vkSj (D) pquus ij +2 vad feysaxs ’ dsoy fodYi (A) pquus ij +1 vad feysaxs ’ dsoy fodYi (B) pquus ij +1 vad feysaxs ’ dsoy fodYi (D) pquus ij +1 vad feysaxs ’ dksbZ Hkh fodYi u pquus ij ¼vFkkZr~ iz'u vuqÙkfjr jgus ij½ 0 vad feysaxs vkSj

vU; fdlh fodYiksa ds la;kstu dks pquus ij �2 vad feysaxs

Paragraph for Question Nos. 1 to 2

iz'u 1 ls 2 ds fy, vuqPNsn

Let 1 1sin (1 {x}).cos (1 {x})

f(x)2{x}.(1 {x})

, where {x} denotes the fractional part of x

ekukfd 1 1sin (1 {x}).cos (1 {x})

f(x)2{x}.(1 {x})

, tgk¡ {x}, x dk fHkUukRed Hkkx Qyu gSA

19. If x 0lim f(x)

is equal to ab (where a & b are co-prime to each other) then

;fn x 0lim f(x)

, ab ds cjkcj gS& (tgk¡ a rFkk b ,d nwljs ds lgvHkkT; gS) rc

(A*) a = 1 (B) a = 2 (C*) b = 2 (D) b = 4

20. x 0

L lim f(x)a b

then which of the following may be CORRECT ?

x 0

L lim f(x)a b

rc fuEu esa ls dkSulk lR; gks ldrk gS ?

(A*) (a, b) (2, 2) (B*) (a, b) (1, 8) (C) (a, b) (1, 4) (D) (a, b) (2, 4) Sol. (5 to 6)

19. 1 1

h 0

sin (1 {0 h}).cos (1 {0 h})lim

2{0 h}.(1 {0 h})

1 1

h 0

sin (1 h) cos (1 h)lim .

(1 h) 2h

In second limit put 1 h cos

nwljh lhek 1 h cos j[kus ij

1 1

h 0 0

sin (1 h) cos coslim limit 2(1 h) 2(1 cos )



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20. 1 1

h 0 h 0x 0

sin (1 {0 h})cos (1 {0 h})lim f(x) lim f(0 h) lim

2{0 h} (1 {0 h})

1 1 1 1

h 0 h 0

sin (1 h 1)cos (1 h 1) sin h cos hlim lim .

h2{0 h} (1 h 1) 2(1 h)

21.2 2 2

Paragraph for Question Nos. 1 to 2 iz'u 1 ls 2 ds fy, vuqPNsn

Consider two functions ekukfd Qyu f(x) = n

n

xlim cos

n

& g(x) = �x4b

Where tgk¡ b = 2 2

xlim x x 1 x 1

, then rc

21. f(x) is gS&

(A) 2xe (B*)

2x2e

(C*) 1

g(x )2e (D) eg(x)

22. Number of solution(s) of equation f(x) + g(x) = 0 is two then can lies in the interval lehdj.k f(x) + g(x) = 0 ds gyksa dh la[;k nks gS] rc dk vUrjky fLFkr gks ldrk gS&

(A) (�, �1) (B) (�7, �2) (C*) (0, ) (D*) (1, 5)

Sol. f(x) =

2 2

n

x xlim

2 2

xsin

2 ne ex

2 n

b = 2 2x

xlim

x x 1 x 1 =

12

(rationalization ifjes;dj.k)

Hence vr% 2x

2e

x2

Total # 2 solutions dqy nks gy



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DATE : 21-06-2020 SET/CODE-1 COURSE : VIJETA (JP) | BATCH : (JPB*, 03JPAB)


PAPER-1

PART Hkkx : II PHYSICS HkkSfrd foKku


This section contains TWELVE (12) questions.

Each question has FOUR options (A), (B), (C) and (D) ONLY ONE of these four option is correct

There are 12 Questions & you have attempt any 9 Questions. If a student attempts more than 9

questions, then only first 9 questions which he has attempted will be checked.

Marking scheme :


Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered).

Negative Marks : �1 In all other cases

[kaM 1 : (vf/kdre vad : 27)

bl [kaM esa ckjg (12) iz'u gSaA



fn;s x;s izFke 9 iz'uksa dh gh tk¡p dh tk;sxhA

vadu ;kstuk :


'kwU; vad % 0 ;fn dksbZ Hkh fodYi ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A


23. Maximum height reached by a rocket fired with a speed equal to 50% of the escape velocity from

earth�s surface is: ¶yk;u osx ds 50% pky ds cjkcj ls izTofyr jkWdsV }kjk i`Foh lrg ls izkIr egRRe Å¡pkbZ gS &

(A) R/2 (B) 16R/9 (C*) R/3 (D) R/8

Sol. V = 50

100Ve =

1 2GM2 R

Apply energy conservation 2GMm 1

m VR 2

= GMm(R h)

v2 = 2GM 2GM

R R h

1 2GM.

4 R =

1 12GM

R R h

1 h

4R R(R h)

R + h = 4h h = R/3


TARGET : JEE (MAIN+ADVANCED) 2021



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24. Distance between the centres of two stars is '5a'. The masses of these stars are M & 81M and their radii are 'a' each. A body of mass � m � is fired straight from the surface of the heavier star towards the

lighter star. What should be its minimum initial speed to reach the surface of the smaller star ? Obtain the expression in terms of G, M & a. (Assume both the stars to be solid spheres with uniform mass distribution)

nks flrkjksa ds dsUnzksa ds chp nwjh '5a' gSA bu flrkjksa ds nzO;eku M o 81M rFkk nksuksa dh f=kT;k a gSA Hkkjhs flrkjs

dh lrg ls � m � nzO;eku ds ,d fi.M dks gYds flrkjs dh vksj Qsadk tkrk gSA gYds flrkjs dh lrg rd igq¡pus ds

fy, bldh U;wure pky fdruh gksuh pkfg,A G, M o a ds inksa esa O;atd çkIr djksA (;g ekfu;s fd nksuksa rkjs ,d

leku Bksl xksys gS)

(A) 675GM

28a (B)

117GMa

(C*) 120GM

a (D) None of these buesa ls dksbZ ugh

Sol. Let gravitational field at P = 0 (P = 0 ij xq:Roh; {ks=k)

2 2

GM 81GM

x (5a x)

x =

a2

Hence there is no null point between x = a to x = 4a. From work energy theorem x = a ls x = 4a ds chp dksbZ mnklhu fcUnq ugh gSA dk;Z ÅtkZ izes; ls

0 � 12

mu2 = GMm 81 GMm 81 GMm GMm

a 4a a 4a

solving gy djus ij u = 120GM

a

x=5a x=0

M 81M

P

x 5a-x

25. A wire of resistance 10 is bent to form a circle. P and Q are points on the circumference of the circle dividing it into a quadrant of 3 V and internal resistance 1 as shown in the figure. The currents in the two parts of the circle are

,d 10 izfrjks/k ds rkj dks eksM+ dj o`Ùk cuk;k x;k gSA P o Q o`Ùk dh ifjf/k ij fcUnq gSa] tksfd ,d 3 V vkSj

vkarfjd izfrjks/k 1 esa fp=kkuqlkj prqFkkZa'k esa foHkkftr djrh gSA o`Ùk ds nksuksa Hkkxksa esa /kkjk gS %

(*A) A236

and A2318

(B) A265

and A2615

(C) A254

and A2512

(D) A253

and A259

26. The resistance R1 is adjusted in such a way that whether the switch is in position 1 or in 2, the current

through the ammeter is I = 2 A. The value of emf of E2 = 6 V. The cells are ideal. The value of resistance R2 is

izfrjks/k R1 dks bl izdkj lek;ksftr fd;k tkrk gS fd pkgs fLop fLFkfr 1 esa gks ;k 2 esa] vehVj ls izokfgr /kkjk I

= 2 A gS E2 ds foHkokUrj dk eku 6 V gSA lsy vkn'kZ gSA izfrjks/k R2 dk eku gSA

A

E1E2

R2

R1

1

2

(A*) Equal to 3 (B) greater than 3 (C) Less than 3 (D) Dependent on E1 (A*) 3 ds cjkcj (B) 3 ls vf/kd (C) 3 ls de (D) E1 ij fuHkZj



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Sol. = 1 1 2

1 1 2

E E ER R R

= 2

2

ER

R2 = 3

27. Which of the following electric field pattern is correct inside the conductor having two materials of

resistivity 1 and 2 (<1) ? izfrjks/kdrk 1 o 2 (<1) okys nks inkFkksZ ls cus ,d pkyd ds vUnj lgh fo|+qr {ks=k izfr:i gSA?

(A*) (B)

(C) (D)

Sol. In series combintim current should be same. Js.kh la;kstu esa /kkjk leku gksuh pkfg,A

E = J Er Ans. 28. A wire has resistance per unit length and is arranged as shown. If side of the bigger square is then

equivalent resistance between points A and B will be ,d rkj dk izfr bdkbZ yEckbZ izfrjks/k gS] vkSj ;g n'kkZ;s vuqlkj O;ofLFkr gSA ;fn cM+s oxZ dh Hkqtk gS rks fcUnq

A o B ds e/; rqY; izfrjks/k gksxk -

/2

A B

(A*)2

(B) 2 (C) 2 (D)

Sol. A B

2

2

2

2

2

2A B

2

2

2

1+ 2

1+ 2

2A B

1+ 2

121 2

= 1 2 1 1

21 2

= 2

2

= 2



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29. A cubical box is used to keep drinks cold in a bus. Surface area of each face is 0.80 m2 and each wall has thickness 2.00 cm. It is filled with ice and water which remains at 0ºC. Temperature outside each

wall is 30ºC then:(Thermal conductivity of material of box=1×10�2w/m�k)(Latent heat of fusion of ice=3.2×105 J/kg)

cl esa is; dks B.Mk j[kus ds fy, ?ku ckWDl ç;qDr gSA çR;sd i`"B dk i`"Bh; {ks=k 0.80 m2 rFkk eksVkbZ 2.00 cm gSA bls cQZ rFkk ty ls Hkjk x;k gS tks 0ºC ij jgrk gSA çR;sd nhokj ds ckgj rkieku 30ºC gS rks : (ckWDl ds inkFkZ

dh Å"eh; pkydrk = 1 × 10�2 w/m�k) (cQZ dh xyu dh xqIr Å"ek = 3.2 × 105 J/kg) (A) Total heat current in the box is 60 J/s (B) Total heat current in the box is 52 J/s (C*) Mass of ice melted in 1 hour is 810 gm (D) Mass of ice melted in 1 hour is 690 gm (A) ckWDl esa dqy Å"eh; /kkjk 60 J/s gSA (B) ckWDl esa dqy Å"eh; /kkjk 72 J/s gSA (C*) 1 ?k.Vs esa fi?kyh cQZ dk nzO;eku 810 gm gSA (D) 1 ?k.Vs esa fi?kyh cQZ dk nzO;eku 690 gm gSA

Sol. H = H CT TKA

L

= (0.01) (0.80). 30º 0º

0.020

H = 12w = 12 J/s So, through all faces = 12 × 6 vr% lHkh i`"Bksa ls = 12 × 6 Htotal = 72 J/sec. = Ht = 72 × 3600 J

m = fL

= 5

72 3600

3.2 10

= 810 gms.

30. A ring consisting of two parts ADB and ACB of same conductivity K carries an amount of heat H. The

ADB part is now replaced with another metal keeping the temperature T1 and T2 constant. The heat

carried increases to 2H. What should be the conductivity of the new ADB part ? ACB

3ADB

,d oy; tksfd leku pkydrk okys nks fgLlksa ADB rFkk ACB ls cus gS dqy Å"ek H dh nj ls çofkgr djrs gSA ADB fgLlk vc ,d nwljs /krq ls cny fn;k tkrk gS ysfdu rkieku T1 rFkk T2 fu;r j[krs gSA vc dqy Å"ek çokg

nj 2H gks tkrh gSA u, ADB fgLls dh pkydrk D;k gksuh pkfg,? ACB

3ADB

C

O

T1 T2 B A

D

(A*) 7

K3

(B) 2K (C) 5

K2

(D) 3K

Sol. In first case :

H = H1 + H2 = 1 2KA(T T )

3

+ 1 2KA(T T )

= 43

1 2KA(T T )

In 2nd case : H1 + H2 = 2H H2 = 2H � H1 [H1 = H1]

H2 = 2H � H1 = 1 28 KA

(T T )3

� 1 2K A

(T T )

= 1 2K A(T T )

K = 73

K



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31. A black body emits radiation at the rate P when its temperature is T. At this temperature the wavelength at which the radiation has maximum intensity is 0. If at another temperature T the power radiated is P

and wavelength at maximum intensity is 0

2

then

rkiØe 'T' ij ,d vkn'kZ df̀".kdk P 'kfä nj ls fofdj.k mRlftZr djrh gSA bl rki ij egRre rhozrk ds

fofdj.k dh rjaxnS/;Z 0 gSA ;fn vU; rki T ij mRlftZr 'kfä P gS rFkk egRre rhozrk dh rjaxnS/;Z 0

2

gS rc &

(A*) P T = 32PT (B) P T = 16PT (C) P T = 8PT (D) P T = 4PT Sol. For a block body, wavelength for maximum intensity : ,d Ñf".kdk oLrq ds fy,] vf/kdre rhozrk ds fy, rjaxnS/;Z

1T

& P T4

4

1P

P = 16 P. P T = 32PT 32. A sphere and a cube of same material and same total surface area are placed in same evacuated

space after they are heated to same temperature. The ratio of their initial rate of cooling in space is : ,d xksyk o ,d ?ku] ,d gh inkFkZ ls cus gS rFkk buds dqy i`"Bh; {ks=k Hkh leku gSA leku rki rd xeZ dj bUgsa

leku fuokZfrr LFkku ij j[kk x;k gSA buds çkjfEHkd 'khryu dh nj dk vuqikr gksxk &

(A) 1 (B) (C*) 6

(D) 1

6

Sol. For sphere : xksys ds fy,

T4 S = m1C. sphere

ddt

For cube : ?ku ds fy,

T4. S = m2.C cube

ddt

sphere

cube

ddt

ddt

= 2

1

mm

= 2

1

VV

[S = 6a2 = (4r2)]

= 6

33. A solid cube of side R and solid sphere of radius R of same material inside room of temperature T0.

If temperature of cube 4T0 and sphere 2T0 then find out ratio of their rates of cooling. T0 rkieku ds dejs ds vUnj j[ks leku inkFkZ ds ,d Bksl ?ku dh Hkqtk R ,oa Bksl xksys dh f=kT;k R gSA ;fn ?ku

dk rki 4T0 ,oa xksys dk rki 2T0 gSA rc muds 'khryu dh nj dk vuqikr Kkr dhft,A

(A*) 34 (B) 32 (C) 12 (D) 28 (E) 278

Sol. 4 40

dT eAC (T T )

dt ms

0

0

4 411 1 24 4

2 2 1 2

T TC A m

C A m T T

32

0 012 3

2 0 0

4R 256T TC 6R 3

C 16T T4 R .R

1

2

C34

C



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34. A rod of uniform cross�section but non�uniform thermal conductivity which vary as k = k0 2

2

xL

(1 x L) (as shown in figure) is kept between fixed temperature difference for a long time. Select the correct option(s) :

,d le:i vuqçLFk dkV {ks=kQy dh ,d NM+ ysfdu vle:i rkih; pkydrk dh ,d NM+ ftldh pkydrk k = k0 2

2

xL

(1 x L) ds vuqlkj ifjofrZr gksrh gSA fp=k eas n'kkZ;s vuqlkj fuf'pr rkikUrj ds e/; yEcs le; ds fy,

j[kh tkrh gSA lgh fodYi@fodYiksa dk p;u dhft, &

(A*) (B) (C) (D) None of these

Sol. i = 2

02

k x dTA

dxL

H

T x2

20T 1

i L dxdT

k A x

T � TH = x2

0 1

i L 1k A x


This section contains SIX (06) questions. The answer to each question is NUMERICAL VALUE with two digit integer and decimal upto two digit.

If the numerical value has more than two decimal places truncate/round-off the value to TWO decimal placed.

Marking scheme : Full Marks : +3 If ONLY the correct option is chosen. Zero Marks : 0 In all other cases

[kaM 2 ¼vf/kdre vad% 18)

bl [kaM esa N% (06) iz'u gSA izR;sd iz'u dk mÙkj la[;kRed eku (NUMERICAL VALUE) gSa]tks f}&vadh; iw.kkZad

rFkk n'keyo f}&vadu eas gSA

;fn la[;kRed eku esa nks ls vf/kd n’'keyo LFkku gS ] rks la[;kRed eku dks n'keyo ds nks LFkkuksa rd VªadsV@jkmaM

vkWQ (truncate/round-off) djsaA

vadu ;kstuk %&

iw.kZ vad % +3 ;fn flQZ lgh la[;kRed eku (Numerical value) gh mÙkj Lo:i ntZ fd;k x;k gSA

'kwU; vad % 0 vU; lHkh ifjfLFkfr;ksa esaA



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35. A planet is made of two materials of density 1 and 2 as shown in figure. ,d xzg ] nks inkFkksZ ls cuk gS ftuds ?kuRo n'kkZ,uqlkj 1 rFkk 2 gS

The acceleration due to gravity at surface of planet is same as a depth �R�. The ratio of 1

2

is

n35.35

. Find the value of n.

xzg dh lrg ij rFkk xgjkbZ �R� ij xq:Roh; Roj.k leku gSA 1

2

vuqikr dk eku

n35.35

gSA n dk eku Kkr

dhft,A Ans. 15.15

Sol. 2

GM

(2R) =

2

GM'

R

M4

= M

43R3 1 +

43(8R3 � R3)2 = 3

14

4 . R .3

1 + 72 = 41

1

2

=

73

36. A man does work W on a ball to throw it vertically upto a height h on earth, if man does work kW on

same ball to throw it vertically upto height 2h on another planet having its radius 14

of radius of earth

and planet has same density as that of earth. Find k251

?

,d O;fDr ,d xsan dks i`Foh ij h Å¡pkbZ rd Å/okZ/kj Qsadus esa] xsan ij W dk;Z djrk gSA ;fn O;fDr mlh xsan dks

vU; xzg ij ftldh f=kT;k] i`Foh dh f=kT;k dh 14 gS rFkk ?kuRo] i`Foh ds ?kuRo ds leku gS] 2h Å¡pkbZ rd

Å/okZ/kj Qsadus esa kW dk;Z djrk gS rks k251

Kkr djks ?

Ans. 12.75 Sol. Wearth = mgeh

= m43

GdR.h

Wplanet = m. 43

Gd.R4

2h

earth

planet

W

W = 2

Wplanet = W2

.



®




37. 300 nos. of identical galvanic cells, each of internal resistance 9 are arranged as several in�series groups of cells connected in parallel. The arrangement has been laid out so that power output in an externally connected resistance of value 16 is maximum. If n number of cells are connected in every

series group that form parallel combination, then find value of 4n

.

300 ,d leku xsYosfud lsy] çR;sd dk vkUrfjd çfrjks/k 9 bl çdkj O;ofLFkr gS fd Js.khØe esa tqM+s lsyksa ds

lewg lekUrj esa tqM+s gSA bl O;oLFkk esa 16 ds ckº; çfrjks/k esa fuxZr 'kfDr vf/kdre gSA ;fn Js.khØe esa tqM+s

lsyksa ds çR;sd lewg esa n lsy gS rks 4n

dk eku Kkr djksA

E1 E2 En

R = 16 Ans. 06.25 Sol. Power developed in it is maximum when req = R

n/N

nr = R (N � total nos. of cells) (n - cell in one series group)

n = r

RN

using values R = 16 , N = 300, r = 9

n =3

40 = 23.1

As n has to be integer and cofactor of 300, then nearest possible values are 25 and 25. 29Power is maximum when current through it is maximum

I =

Nrn

R

En2

using n = 25 and n = 20 and solving we get 420

E300,

417E300

2025

It is obvious that when n = 25, I is greater than when n = 20. Therefore n = 25 nos. Sol. tc mRiUu 'kfDr vf/kdre gS req = R

n/N

nr = R (N � dqy lSyksa dh la[;k)

(n - ,d Js.kh lewg esa lSyksa dh la[;k)

n = r

RN

ekuksa dk iz;ksx djus ij R = 16 W, N = 300, r = 9 W

n = 3

40 = 23.1

pwafd n iw.kk±d gS rFkk ;g 300 dk xq.kd gS rks laHkkfor lehire eku 20 rFkk 25 gS 'kfDr vf/kdre gksxh tc izokfgr /kkjk vf/kdre gksA

I =

Nrn

R

En2

n = 25 o n = 20 j[kus ij ,oa gy djus ij

420

E300,

417E300

2025

vFkkZr~ tc n = 25 gS rks , n = 20 dh vis{kk vf/kdre gS vr% n = 25 gksxk



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38. A cell of emf 1.5 V and internal resistance 0.5 is connected to a non-linear conductor in which current varies with voltage as V2 = . If the current drawn from cell is 'I', then find 12.75×I(in ampere).

1.5 V fo-ok-cy rFkk 0.5 vkUrfjd çfrjks/k ds lsy dks vjs[kh; pkyd ls tksM+k x;k gS] ftlesa /kkjk, oksYVst ds

lkFk V2 = vuqlkj ifjorhZ gSA ;fn lsy ls yh xbZ /kkjk 'I' gS, rks 12.75×I (,sfEi;j ek=kd esa) Kkr dhft,A Ans. 12.75

Sol.

From conservation of energy ÅtkZ laj{k.k ls E � V = r E � V = V2 r V2 r + V � E = 0

1 4Er 1

V2r

V = 1 V 1.5 � 1 = × 0.5 = 1 A 39. A body of mass 4 kg is suspended from a spring of spring constant 400 N/m. Another body of mass

4 kg moving vertically upward with 2 m/s hits it & gets embedded in it. If amptitude is 1 3x 4

m. find

'5.05x'. nzO;eku 4 kg dh ,d oLrq 400 N/m fLçax fu;rkad ds ,d fLçax ls yVdh gSA 4 kg nzO;eku dh ,d vU; oLrq

2 m/s ls Åij dh xfr'khy gS ,oa blls Vdjkdj blesa ?kql tkrh gSA ;fn vk;ke 1 3x 4

m gS rks '5.05x' Kkr

dhft,A

Ans. 25.25

Sol. = Km

= 400

508

2 = 50

V = 2 2A � x V2 = 2(A2 � x2)

1 = 50 2 1A �

100

A2 = 1 1

50 100 , A2 =

3100

A = 1 35 4

x = 5



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40. A pendulum has period 51s for small oscillations. An obstacle is placed directly beneath the pivot, so that only the lowest one quarter of the string can follow the pendulum bob when it swings to the left of its resting position as shown in the figure. The pendulum is released from rest at a certain point A. The time taken by it to return to that point is X s. Find the minimum value of X ?

,d yksyd ds fy;s vYi nksyuksa dk vkorZdky 51s gSA ,d vojks/kd dks dhyd ds Bhd uhps yxk fn;k tkrk gS

rkfd fojkekoLFkk ls NksM+us ij tc yksyd dk xksyd viuh ek/; fLFkfr ls xqtjs rks blds ckn og ewy yEckbZ dh

,d pkSFkkbZ yEckbZ ls gh fp=kkuqlkj cka;h vksj nksyu djsA yksyd dks fcUnq A ls fojkekoLFkk ls NksM+k tkrk gSA bls

okil blh fcUnq rd iqu% ykSVus esa yxk le; Xs gSA X dk U;wure~ eku Kkr djks ?

Ans. 38.75

Sol. T length free length becomes 4

So time period becomes T1 = T2

T yEckbZ eqDr yEckbZ 4

gks tk;sxhA vr% vkorZdky T1 = T2

gksxk

Half oscillation is with free length and half with free length 4

vk/ks nksyu esa Mksjh dh yEckbZ gS rFkk 'ks"k vk/kk nksyu 4

ds lkFk gSA

Total time dqy le; = T2

+ 1T

2 =

3T4










®



















ç'u 41 ls 42 ds fy, vuqPNsn

A point source of sound 'S' generating sound of 875 Hz is moving along straight line with speed 50 m/s as shown in figure. A stationary observer 'O' is situated at a perpendicular distance 300 m from CD as shown. Speed of sound is 300 m/s. Based on given information answer the following :

,d fcUnqor~ /ofu L=kksr 'S' ljy js[kk ds vufn'k 50 m/s pky ls xfr djrs le; fp=kkuqlkj 875 Hz vko`fÙk dh

/ofu mRiUu djrk gSA fp=kkuqlkj CD js[kk ls 300 m yEcor~ nwjh ij ,d fLFkj çs{kd 'O' fLFkr gSA /ofu dh pky

300 m/s gSA mijksDr lwpuk ds vk/kkj ij fuEu dk mÙkj nhft, :

41. Calculate frequency of sound observed by observer when source is at point A : tc L=kksr fcUnq A ij gS rc çs{kd }kjk çsf{kr vko`fÙk Kkr djks & (A*) 900 Hz (B) 899 Hz (C) 875 Hz (D) 600 Hz Sol.

f' = 2

2 2

cf

c u =

2

2 2

300875

300 50

= 900 Hz.



®




42. Distance between observer and source when observer hears original frequency : tc çs{kd okLrfod vkof̀Ùk lwurk gS rc L=kksr rFkk çs{kd ds e/; nwjh Kkr djks & (A) 300 m (B) 305 m

(C*) 50 37 m (D) None of these buesa ls dksbZ ugha Sol.

OS = 2 2300 50 = 925 = 50 37 m.

Paragraph for Question Nos. 43 to 44 ç'u 43 ls 44 ds fy, vuqPNsn

A string fixed at both the ends of length 2 m, vibrating in its 7th overtone. Equation of the standing wave is given by y = A sin kx cos (t + /3), All the symbols have their usual meaning. Mass per unit length of the string is 0.5 gm/cm. Given that A = 1 cm and = 100 rad/sec. Answer the following 2 questions based on information given (Use 2 = 10)

2 m yEckbZ dh jLlh nksuksa fljksa ij tM+or~ cU/kh gqbZ gS, ;g 7 osa vf/kLojd esa dEiUu dj jgh gSA blesa mRiUu

vizxkeh rjax dh lehdj.k y = A sin kx cos (t + /3) }kjk nh tkrh gSA ;gk¡ lHkh izrhdksa ds lkekU; vFkZ gSA

jLlh dh bdkbZ yEckbZ dk nzO;eku 0.5 gm/cm gSA fn;k gS A = 1 cm rFkk

= 100 rad/secA nh xbZ lqpuk ij vk/kkfjr fuEu nks iz'uksa ds mÙkj nhft,A (2 = 10) 43. Staring from t = 0, energy of vibaration is completely potential at time t, where t is : t = 0 ls izkjEHk djrs gw,] fdlh le; t ij nkSyu ÅtkZ iw.kZr% fLFkfrt ÅtkZ ifjofrZr gks tkrh gSA tgk¡ t gksxkA

(A*) 1

150 sec. (B*)

160

sec. (C) 3

100 sec (D*)

11300

sec.

44. Starting from t = 0, energy of vibration is completely kinetic at time t, where t is :

t = 0 ls izkjEHk djrs gw,] fdlh le; t ij nkSyu ÅtkZ iw.kZr% xfrt ÅtkZ ifjofrZr gks tkrh gSA tgk¡ t gksxkA

(A*) 1

600 sec. (B)

5600

(C*) 19600

(D*) 25

6000 sec.

Sol. for energy to be completely potential ÅtkZ ds iw.kZ :i ls fLFkfrt ÅtkZ ifjofrZr gksus ds fy,

cos t3

= + 1 t3

= n

t = 3n 1

3

=

3n 1300

sec.

for energy to be completely kinetic ÅtkZ ds iw.kZ :i ls xfrt ÅtkZ ifjofrZr gksus ds fy,

cos t3

= 0

t + 3

= (2n � 1) 2

t = 6n 5600

sec.



®





TARGET : JEE (MAIN+ADVANCED) 2021DATE : 19-07-2020 SET/CODE-1 COURSE : VIJETA (JP) | BATCH : (JPB*, 03JPAB)


PAPER-1 PART : III CHEMISTRY


This section contains TWELVE (12) questions. Each question has FOUR options (A), (B), (C) and (D) ONLY ONE of these four option is correct There are 12 Questions & you have attempt any 9 Questions. If a student attempts more than 9

questions, then only first 9 questions which he has attempted will be checked. Marking scheme :

Full Marks : +3 If ONLY the correct option is chosen. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : �1 In all other cases

[kaM 1 : (vf/kdre vad : 27) bl [kaM esa ckjg (12) iz'u gSaA



fn;s x;s izFke 9 iz'uksa dh gh tk¡p dh tk;sxhA vadu ;kstuk :


'kwU; vad % 0 ;fn dksbZ Hkh fodYi ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A _.k vad % �1 vU; lHkh ifjfLFkfr;ksa esaA

45. In which of the following electrolysis in aqueous medium, mass of anode and pH of solution both

remains unchanged? (A) Electrolysis of aqueous AgNO3 using silver anode and copper cathode. (B) Electrolysis of aqueous CuSO4 using pure copper anode and impure copper cathode. (C) Electrolysis of aqueous CuSO4 using silver anode and platinum cathode. (D*) Electrolysis of aqueous AgNO3 using gold anode and silver cathode.

tyh; ek/;e esa fuEu esa ls dkSulk oS|qr vi?kVu esa ,uksM dk æO;eku rFkk foy;u dh pH nksuksa vifjofrZr jgrh

gS ?

(A) flYoj ,uksM rFkk dkWij dSFkksM+ dk iz;ksx djrs gq, tyh; AgNO3 dk oS|qr vi?kVu (B) 'kq) dkWij ,uksM rFkk v'kq) dkWij dSFkksM+ dk iz;ksx djrs gq, tyh; CuSO4 dk oS|qr vi?kVu (C) flYoj ,uksM rFkk IysfVue dSFkksM+ dk iz;ksx djrs gq, tyh; CuSO4 dk oS|qr vi?kVu (D*) xksYM ,uksM rFkk flYoj dSFkksM dk iz;ksx djrs gq, tyh; AgNO3 dk oS|qr vi?kVu Sol. Anode Cathode (A) Ag Ag+ + e� Ag+ + e� Ag

(B) Cu Cu2+ + 2e� Cu2+ + 2e� Cu

(C) Ag Ag+ + e� Ag+ + e� Ag

(D) 2H2O O2 + 4H+ + 4e� Ag+ + e� Ag Sol. ,uksM dSFkksM

(A) Ag Ag+ + e� Ag+ + e� Ag

(B) Cu Cu2+ + 2e� Cu2+ + 2e� Cu

(C) Ag Ag+ + e� Ag+ + e� Ag

(D) 2H2O O2 + 4H+ + 4e� Ag+ + e� Ag



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46. Orbital angular momentum of an electron is

h3 , then the number of possible orientations of this

orbital in space is:

,d bysDVªkWu dk d{kd dks.kh; laosx

h3 gS rks vUrfj{k esa bl d{kd ds laHko foU;klksa dh la[;k gksxh &

(A) 3 (B) 5 (C*) 7 (D) 9 Sol. Orbital angular momentum of electron

=

h

.32h

.)1l(l

l = 3 number of orientation = 2l + 1 = 2 × 3 + 1 = 7 Sol. bysDVªkWu dk d{kd dks.kh; laosx

=

h

.32h

.)1l(l

l = 3 foU;kl dh la[;k = 2l + 1 = 2 × 3 + 1 = 7 47. Which of the following will not produce a precipitate with Ba(OH)2? (A) Chromyl chloride (B) CO2 (C) SO2 (D*) NaNO2

fuEu es ls dkSulk Ba(OH)2 ds lkFk vo{ksi mRikfnr ugha djsxk\ (A) Øksfey DyksjkbM (B) CO2 (C) SO2 (D*) NaNO2

48. Which of the following statement about Ellingham diagram is correct ?

(A) On increasing temperature, stability of metal oxides increases (B*) Reduction of metal oxide is easier if the metal formed is in gaseous state than in solid state at the temperature of reduction (C) Reduction of metal oxide is easier if the metal formed is in solid state than gaseous state at the temperature of reduction (D) Suitable reducing agent for metal oxide reduction is predicted only on the basis of enthalpy change of reaction

,fyaxe vkjs[k ds lanHkZ esa fuEu esa ls dkSulk dFku lgh gS\ (A) rkieku c<+us ij /kkrq vkWDlkbM dk LFkkf;Ro c<+rk gSA (B*) /kkrq vkWDlkbM dk vip;u vklku gksrk gS ;fn fufeZr /kkrq vip;u rki ij Bksl voLFkk dh ctk;s xSlh;

voLFkk esa gksrh gSA (C) /kkrq vkWDlkbM dk vip;u vklku gksrk gS ;fn fufeZr /kkrq vip;u rki ij xSlh; voLFkk dh ctk;s Bksl

voLFkk esa gksrh gSA (D) /kkrq vkWDlkbM ds vip;u ds fy, mfpr vipk;d vfHkdeZd dsoy vfHkfØ;k dh ,UFkSYih ifjorZau ds vk/kkj ij

crk;k tkrk gSA

Sol. Slope is positive so T↑, ∆G become less �ve and stability of metal oxides decreases. In liquid state slope is more positive that�s why reduction of metal oxide become more easier.

Suitable reducing agent for metal oxide reduction is predicted on the basis of ∆G of reaction. Sol. <+ky /kukRed gksrk gS vr% T↑, ∆G de _.kkRed gks tkrk gS rFkk /kkrq vkWDlkbM dk LFkkf;Ro ?kVrk gSA nzo voLFkk esa <+ky vf/kd /kukRed gksrk gS ftlls /kkrq vkWDlkbM dk vip;u vf/kd vklkuh ls gksxkA 49. Which of the following aq. solution turn 'milky' with CO2? fuEu es ls dkSulk tyh; foy;u] CO2 ds lkFk nwf/k;k gks tkrk gS\ (A) CaCl2 (B) NaOH (C) AgNO3 (D*) Ba(OH)2

50. Wolframite is separated from cassiterite by : (A) froth floatation method (B) levigation (C*) electromagnetic separation method (D) electrostatic separation method okWYÝsekbV dks dSlhVsjkbV ls fuEu }kjk i`Fkd fd;k tkrk gSa& (A) >kx Iyou fof/k }kjk (B) /kkou }kjk (levigation) (C*) fo|qr pqEcdh; i`Fkd~dj.k fof/k }kjk (D) fLFkj oS|qr i`Fkd~dj.k fof/k }kjk

Sol. Wolframite is ferromagnetic in nature and, therefore, it is separated from cassiterite ore (non-magnetic) by electromagnetic separation method.



®




Sol. okWYÝsekbV ykSgpqEcdh; izd`fr dk gS blfy, dSlhVsjkbV v;Ld ¼vpqEcdh;½ ls bldk i`FkDdj.k oS|qrpqEcdh;

i`FkDdj.k fof/k }kjk fd;k tkrk gSA 51. Using the data given below find the incorrect statement regarding for the given species : o

Cr/OCr 3�272

E = 1.33 V oCl/Cl �

2E = 1.36 V

oMn/MnO 2�

4E = 1.51 V o

Cr/Cr3E = � 0.74 V

(A) Strongest reducing agent is Cr amongst Cl�, Mn2+, Cr & Cr3+

(B) Strongest oxidising agent is �4MnO amongst Cl�, Cl2, �

4MnO & Cr3+ (C) Order of reducing power is Mn2+ < Cl� < Cr3+ < Cr. (D*) Most stable in its reduced form is Cr amongst Cl�, Cr3+, Cr and Mn2+. uhps fn;s x;s vk¡dM+ks dk iz;ksx djrs gq, nh xbZ Lih'kht ds fy, xyr dFku Kkr dhft, &

oCr/OCr 3�2

72E = 1.33 V o

Cl/Cl �2

E = 1.36 V

oMn/MnO 2�

4E = 1.51 V o

Cr/Cr3E = � 0.74 V

(A) Cl�, Mn2+, Cr rFkk Cr3+ esa ls izcyre vipk;d Cr gSA

(B) Cl�, Cl2, �4MnO rFkk Cr3+ esa ls izcyre vkWDlhdkjd

�4MnO gSA

(C) vipk;d {kerk dk Øe gS Mn2+ < Cl� < Cr3+ < Cr (D*) Cl�, Cr3+, Cr rFkk Mn2+ esa ls buds vipf;r :i es lokZf/kd LFkk;h Cr gSA Sol. Most stable in its reduced form is Mn2+ amongs Cl�, Cr3+, Cr and Mn2+.

Cl�, Cr3+, Cr rFkk Mn2+ esa ls buds vipf;r :i es lokZf/kd LFkk;h Mn2+ gSA 52. Select correct statement from the following :

(A) is more basic than (B) is more acidic than

(C*) HCCH is more acidic than NH3 (D) is more stable than

lgh dFku dk p;u dhft,A

(A) dh vis{kk vf/kd {kkjh; gksrk gSA

(B)

dh vis{kk vf/kd vEyh; gksrk gSA

(C*) NH3 dh vis{kk HCCH vf/kd vEyh; gksrk gSA

(D) dh vis{kk vf/kd LFkk;h gksrk gSA

Sol. Self explanatory. Lor% le>us ;ksX;A 53. The compound which has maximum number of chiral centres is fuEu ;kSfxdksa esa ls dkSuls ;kSfxd esa fdjSy dkcZu dh la[;k vf/kdre gS %

(A) (B) (C*) (D)



®




Sol. Four chiral carbon atoms (pkj fdjSy dkcZu ijek.kq)

54. 2

(1) excess PhMgBr(2) H O

X , X is gS

(A) (B) (C*) (D)

Sol. PhMgBr

PhOMgBr H O2

PhOH

PhOH

55. Give the decreasing order of nucleophilic addition reaction of the following : (i) HCHO (ii) PhCHO (iii) Chloral (Cl3C�CH=O) (iv) Acetophenone fuEu ;kSfxdksa esa ukfHkdLusgh ;ksxkRed vfHkfØ;k ds fy, ?kVrh gqbZ nj dk Øe gS % (i) HCHO (ii) PhCHO (iii) Dyksjy (Cl3C�CH=O) (iv) ,lhVksfQuksu (A) iii > i > ii > iv (B) iv > ii > i > iii (C*) i > iii > ii > iv (D) iii > i > iv > ii Sol. Reactivity of Neucleophilic addition reaction EWG. Sol. ukfHkdLusgh ;ksxkRed vfHkfØ;k dh nj EWG ds lekuqikrh gksrh gSA

56. Predict the product formed in the following reaction fuEu vfHkfØ;k esa fufeZr mRikn crkb;s&

mRikn

(A) (B*) (C) (D)

SECTION � 2 : (Maximum Marks : 18) This section contains SIX (06) questions. The answer to each question is NUMERICAL VALUE with two

digit integer and decimal upto two digit. If the numerical value has more than two decimal places truncate/round-off the value to TWO decimal

placed. Marking scheme :


Zero Marks : 0 In all other cases

[kaM 2 ¼vf/kdre vad% 18) bl [kaM esa N% (06) iz'u gSA izR;sd iz'u dk mÙkj la[;kRed eku (NUMERICAL VALUE) gSa]tks f}&vadh; iw.kkZad

rFkk n'keyo f}&vadu eas gSA ;fn la[;kRed eku esa nks ls vf/kd n’'keyo LFkku gS ] rks la[;kRed eku dks n'keyo ds nks LFkkuksa rd VªadsV@jkmaM

vkWQ (truncate/round-off) djsaA vadu ;kstuk %&

iw.kZ vad % +3 ;fn flQZ lgh la[;kRed eku (Numerical value) gh mÙkj Lo:i ntZ fd;k x;k gSA 'kwU; vad % 0 vU; lHkh ifjfLFkfr;ksa esaA



®




57. How many of following metals can be extracted by self reduction method? fuEu es ls dqy fdrus /kkrqvksa dks Lor% vip;u fof/k }kjk fu"df"kZr fd;k tk ldrk gSA Ca, Sn, Sb, Hg, Cu, Zn, Pb, Al , Na Ans. 04.00 Sol. Sb, Hg, Cu & Pb can be extracted by self reduction method Sb, Hg, Cu o Pb dks Lor% vip;u fof/k }kjk fu"df"kZr fd;k tk ldrk gSA 58. Consider the following molecules: O2, Cl2, Li2, H2, B2, C2 and N2 Let a = number of molecules having (bond order = 1) b = number of molecules having (bond order = 2) c = number of molecules having (bond order = 3) d = number of paramagnetic molecules. Find the value of (a × b × c × d × 4.2).

fuEu v.kqvks ij fopkj dhft, : O2, Cl2, Li2, H2, B2, C2 rFkk N2

ekuk

a = (ca/k Øe = 1) j[kus okys v.kqvks dh la[;k b = (ca/k Øe = 2) j[kus okys v.kqvks dh la[;k

c = (ca/k Øe = 3) j[kus okys v.kqvks dh la[;k d = vuqpqEcdh; v.kqvks dh la[;k] (a × b × c × d × 4.2) Kkr dhft,A

Ans. 67.20 Sol. a = Cl2, Li2, H2, B2 = 4 b = O2, C2 = 2 c = N2 = 1

d = O2, B2 = 2 59. Evolution of a gas or acid vapours on addition of concentrated H2SO4 to an aqueous solution may

indicate presence of how many of the following acidic radicals in the solution? Chloride, Carbonate, Bromide, Nitrite, Nitrate, Sulphite, Sulphide, Sulphate, Acetate, Phosphate ,d tyh; foy;u esa lkafnzr H2SO4 feykus ij xSl ;k vEy ok"iksa dk fu"dklu foy;u esa fuEu esa ls fdrus

vEyh; ewydksa dh mifLFkfr dks bafaxr dj ldrk gS\ DyksjkbM, dkcksZusV, czksekbM, ukbVªkbV, ukbVªsV, lYQkbV, lYQkbM, lYQsV, ,lhVsV, QkWLQsV Ans. 08.00 Sol. Acidic radical which evolve gases on treatment with dilute HCl also show evolution of gas with

concentrated sulphuric acid. Except sulphate and phosphate all the given radicals show evolution of gas or acid vapours on treatment with conc. H2SO4.

Sol. vEyh; ewyd tks ruq HCl ds lkFk mipkj.k ij xSl fu"dkflr djrk gS] lkafnzr lY¶;wfjd vEy ds lkFk Hkh xSl dk

fu"dklu n'kkZrk gSA lYQsV rFkk QkWLQsV dks NksM+dj fn;s x;s lHkh ewyd lkanz H2SO4 ds lkFk mipkfjr djus ij

xSl ;k vEy ok"i dk fu"dklu n'kkZrs gSaA 60. Calculate acid dissociation constant for 0.1 M HCOOH if its solution shows a resistance of 50 when

filled in a cell having separation between parallel electrodes 4 cm and cross section area of electrode 10 cm2

Given, ])HCOO(Ca[ 2m

= 230 Scm2 mol�1

280]CaCl[ 2m Scm2 mol�1

]HCl[m

= 425 Scm2 mol�1

(Fill your answer by multiplying the answer with 103) 0.1 M HCOOH ds fy, vEy fo;kstu fLFkjkad ifjdfyr dhft;s ;fn bldk foy;u 50 dk izfrjks/k n'kkZrk gS

tc lekUrj bysDVªkWMks ds e/; 4 cm nwjh j[kus okys ,d lsy esa bls Hkjk tkrk gS rFkk bysDVªkWM dk vuqizLFk dkV

{ks=kQy 10 cm2 gSA

fn;k gS ])HCOO(Ca[ 2m

= 230 Scm2 mol�1

280]CaCl[ 2m Scm2 mol�1

]HCl[m

= 425 Scm2 mol�1

(viuk mÙkj 103 ls xq.kk djus ds i'pkr~ nhft;s)



®




Ans. 05.00

Sol. K = 1 1 4R a 50 10

= 8 × 10�3 Scm�1

1.0

1000Km

= 80 S cm2 mol�1

m (HCOOH) = )CaCl(

21

)HCOO(Ca21

)HCl( 22m

= 425 + 115 � 140 = 400 S cm2 mol�1

2.0m

m

Ka =

11.0 2

= 8.0

04.01.0 = 5 × 10�3 M

61. What is the maximum number of moles of CH3MgCl that can be consumed by one mole of phosgene ? CH3MgCl ds eksykss dh vf/kdre la[;k D;k gSa tks 1 eksy QksLthu ls vfHkfØ;k djrs gSa \ Ans. 03.00

Sol.

62. How many of the following substituents present on benzene ring can cause aromatic electrophilic

substitution faster than benzene? fuEu çfrLFkkfi;ksa esa ls fdrus csUthu dh rqyuk esa ,sjksesfVd bysDVªkWu Lusgh izfrLFkkiu vfHkfØ;k csUthu ij rhoz

nsrk gSA

(a) �NH2 (b) �NR2 (c) �NO2 (d) 3�NH

(e)

O||

�O � C �R (f)

O||

�NH �C �R (g)

O||�C � Cl (h)

O||�C �H

(i) �SO3H (j) �CH3 (k) �CH = CR2

Sol. 06.00 (a , b , e, f , j , k)










®



















iz'u 63 ls 64 ds fy, vuqPNsn

A white solid (A) was dissolved in dil.HNO3 to liberate a reddish-brown gas (B) to obtain a colorless solution (C). (B) when absorbed in KOH, gave a colorless solution of (D) and (E). Treatment of (D) with AgNO3 precipitated (A) and gave a solution of (E).

,d lQsn Bksl (A) ruq HNO3 es ?kqydj yky&Hkwjh xSl (B) fu"dkflr djrk gS rFkk ,d jaxghu foy;u (C) izkIr gksrk gSA (B) tc KOH es vo'kksf"kr gksrk gS rks (D) rFkk (E) dk jaxghu foy;u nsrk gSA (D) dks AgNO3 ds lkFk

mipkfjr djus ij ;g (A) dk vo{ksi rFkk foy;u (E) nsrk gSA 63. Which of the following contain at least one non-metal in its highest possible oxidation state? fuEu es ls fdles de ls de ,d v/kkrq] bldh mPpre lEHko vkWDlhdj.k voLFkk es mifLFkr gS\ (A) A (B) B (C*) C (D) D 64. Solution (C) will not give a precipitate with excess of which of the following? (A) Borax (B*) NH3 (C) dil. HCl (D) KOH foy;u (C) fuEu es ls fdlds vkf/kD; ds lkFk vo{ksi ugha nsxk\

(A) cksjsDl (B*) NH3 (C) ruq HCl (D) KOH

Paragraph for Question Nos. 65 to 66 iz'u 65 ls 66 ds fy, vuqPNsn

Nucleophilic aliphatic substitution reaction is given by those compounds which have electron rich groups as leaving groups. Less is the basicity of the leaving group, more is its leaving power.

R � L + Nu

R � Nu + L



®




In the given reaction, L is the leaving group which leaves as nucleophile. Nu

is the incoming group which is always nucleophilic in character. The reaction is nucleophilic substitution reaction which can be unimolecular or bimolecular reaction.

ukfHkdLusgh ,yhQsfVd izfrLFkkiu vfHkfØ;k mu ;kSfxd }kjk nh tkrh gS ftuesa bysDVªkWu /kuh lewg fu"dkflr lewg

ds :i esa gksA fu"dkflr lewg dh {kkjdrk ftruh de gksrh gS mruh gh vf/kd fu"dklu 'kfDr gksrh gSA

R � L + Nu

R � Nu + L

bl vfHkfØ;k esa] L ,d fu"dkflr lewg gS tks fd ukfHkdLusgh ds :i esa”fu"dkflr gksrk gSA Nu

,d vkus okyk lewg

gS tks fd lnSo UkkfHkdLusgh gksrk gSA vfHkfØ;k ,d ukfHkdLusgh izfrLFkkiu vfHkfØ;k gS tks fd ,dk.kwd ;k f}v.kwd

vfHkfØ;k gks ldrh gSA

65. Leaving power of which group is maximum ? fuEu esa ls fdldh fu"dklu lewg lkEFk;Z lokZf/kd gSA

(A) ” (B*) (C) (D) �Cl

Sol. Leaving group in which the negative charge or lone pair of electrons is stabilized, is a good leaving group.

og fu"dklu lewg vPNk gksrk gS] ftldk _.kkos'k ;k ,dkdh bySDVªkWu ;qXe LFkkuhd`r gksA

66. Which one of the following is strong base but poor nucleophile ? fuEu esa dkSu izcy {kkj gS fdUrq nqcZy ukfHkdLusgh gSA

(A) 3CH

(B) HO

(C) (D*)

Sol. (CH3)3 CO

is strong base due to +I effect of three �CH3 group but poor Nu due to steric hinderance.

(CH3)3 CO

,d izcy {kkj gS] rhu �CH3 lewg ds +I izHkko ds dkj.k ijUrq mudh f=kfoe ck/kk ds dkj.k nqcZy

ukfHkdLusgh gSA



QUESTIONS, HINTS & SOLUTIONS MATHEMATICS PAPER-1 · 2020. 11. 28. · PAPER-1 SECTION Œ 1 :...

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Transcript of QUESTIONS, HINTS & SOLUTIONS MATHEMATICS PAPER-1 · 2020. 11. 28. · PAPER-1 SECTION Œ 1 :...